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View Full Version : Development Of An Aeroelastic Stability Boundary Of A Rotor In Autorotation

kolibri282
02-06-2012, 11:31 AM
author: J. Trchalik, E. A. Gillies, D. G. Thompson
comment: The blade torque distribution in fig. 4 seems to give a somewhat different picture than usual.
http://eprints.gla.ac.uk/4967/1/Development_of_an_aeroelastic_stability_boundary_o f_a_rotor_in_autorotation.pdf

Rotor Rooter
02-06-2012, 02:33 PM
Juergen, the following is only speculation and I am not sure if it is answering your concern.

Fig. 4 appears to be a more elaborate version of Leishman's fig. 5. Both diagrams are displaying the torque, unlike the more popular thrust distribution diagrams. An associated thrust disk would show a bilateral thrust, or to be more precise 'a bilateral moment of thrust'.

In other words fig. 4 and 5 are showing the power/drag vectors not the lift vectors.

Dave

kolibri282
02-07-2012, 01:37 PM
Dave,

since the caption says that the torque is charted in fig 4/5 my idea is that the rotor force component laying in the tip path plane is displayed (times the rotor station). If the component is in the direction of rotation it's positive (driving) whereas if it's negative it is driven. What struck me is the fact that on the retreating side there is almost no driven region. From about 135° there is a small driven region close to the blade root and that's it. Leishman's diagram shows a driven region of at least 20% radius over the outer blade which seemed to be the accepted theory so far but which does not exist according to Trchalik's calculations. Very interesting.

Rotor Rooter
02-07-2012, 03:26 PM
What struck me is the fact that on the retreating side there is almost no driven region. From about 135° there is a small driven region close to the blade root and that's it. Leishman's diagram shows a driven region of at least 20% radius over the outer blade which seemed to be the accepted theory so far but which does not exist according to Trchalik's calculations. Very interesting.

It is interesting. Maybe Trchalik's calculations are more exact.

Sikorsky's original ABC (http://www.unicopter.com/0891.html)was limited in forward speed due to the high rotational speed that the rotors were being taken to by autorotation, and Chuck B. has commented on the forward speed limitation of gyrocopters.

Perhaps this also plays a limiting role in the CarterCopter. http://www.unicopter.com/NoIdea.gif

Dave

L_Butler
02-08-2012, 02:29 PM
Juergen,
I found figure 14 to be most interesting. However I'm confused with the axis titles. Kcrit is in units of N-M-M/rad. What is the extra M? Torsional rigidity is N-M/rad.

Also do you know what the Ycc units are? I'm guessing it's percentage of chord. That is, the CG displacement behind the neutral axis as a ratio of the chord.

This appears to confirm C. Beaty's comments on tail heavy rotors and explains why nose weights were so prevalent on early wooden blades.

Thanks,
Larry

kolibri282
02-08-2012, 03:07 PM
Larry,

you guessed correctly that Ycc is percentage of chord, so it has no dimension (i.e. the dimension is 1 as the term in brackets states). The Newton meter squared is because you want to get the angle of twist (in rad) when you apply a moment (having dimensions of Newton * meter) per meter of length of the blade. If you apply a moment to a section that is two meters in length the twist angle is twice that of the same moment acting on a section of one meter, so you have to divide by length of the section. That's what the second meter in Nm*(m) - or Nm^2 - is for. (Hope that makes sense, if not just ask... I'll try a better answer tomorrow... it's now eight minutes past midnight over here and I'm fading fast...;-)

Cheers,

Juergen

L_Butler
02-08-2012, 07:35 PM
Juergen,
I just love the different terminology used throughout the world. I thought the 1 in brackets was a reference to: Note 1. I was looking all over for notes and couldn't find any.

However studying the graph a little more I don't think my assumption of the CG behind the neutral axis is correct. The verbiage above the graph indicates that the neutral axis is at 32% chord and the CG is at 40% chord. Do you know what the Ycg is a measure of? I can't find it in the report.

I'm still scratching my head about the extra M. If I want to determine the twist angle I multiply the length times K. The K factor is a property of the material and geometry. Obviously the greater the K the less torsional deflection seen by the object.

I thought the graph was putting some numbers to the relationship between CG location and torsional rigidity, quantifying where aerodynamic instability occurs. Now I'm confused by the parameters being used.

Vance, I feel your pain.

Thanks,
Larry

kolibri282
02-09-2012, 02:45 AM
The tiwst angle is

fhi = Ks * Moment * Length

If you state it that way the angle will increase as either the moment or the length of
the part or both increase. Here Ks depends on the shape of the section and the material
you use. A moment has dimensions of force times lever arm, which is a length, so it's
[Nm] (square brackets are often used to indicate that this is a dimension). The length
of the part also has a dimension of meters so the combined dimension is

[Nm]*[m] = Nm^2 (Newton meter squared).

Since an angle has no dimension Ks must have dimensions of one divided by Nm^2.

Ks is 1/(G*I) where G is the modulus of rigditiy of the material (dimension = [N/m^2] =
newton per square meter) and Ip is the polar second area moment of the section
which has dimensions of m^4 (meter to the power of four). If you multiply the two
one square meter cancels leaving [Nm^2] in the denominator as is required by the
need for dimensional consistency in any formula.

G*I is the torsional stiffness of a blade taking into account both material and section

As for YCG I think that this is the location of the center of mass. The aerodynamic
center is at 32% chord, so if the CoM is at 32% as well a low torsional stiffness
(actually zero) is required to avoid flutter. The further the CoM moves away from
the aerodynamic center the higher the torsional stiffness has to be to avoid flutter.
That is - from my understanding - what fig. 14 depicts.

Cheers,

Juergen

L_Butler
02-09-2012, 10:42 AM
Juergen,
OK thanks, I was trying to make it more difficult than it really was.

I think the CG verbiage accompanying the graph is an error or incomplete. The body of the report discusses moving the CG, so the 40% reference doesn't belong with the graph. Or is should also mention that calculations were also done at 32%, 36% and 46%, which would match the data points.
Larry